Question: $\dfrac{1 + \cot^2\theta}{\cot^2\theta} = \; ?$
We can derive a useful identity from ${\sin^2 \theta} + {\cos^2 \theta} = 1$ to simplify this expression. $1$ ${\sin\theta}$ ${\cos\theta}$ $\theta$ We can see why this identity is true by using the Pythagorean Theorem. Dividing both sides by $\sin^2\theta$ , we get $ \dfrac{\sin^2\theta}{\sin^2\theta} + \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{1}{\sin^2\theta}$ $ 1 + \cot^2\theta = \csc^2\theta$ Plugging into our expression, we get $ \dfrac{1 + \cot^2\theta}{\cot^2\theta} = \dfrac{\csc^2\theta}{\cot^2\theta} $ To make simplifying easier, let's put everything in terms of $\sin$ and $\cos$ . We know $\csc^2\theta = \frac{1}{\sin^2\theta}$ and $\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta}$ , so we can substitute to get $ \dfrac{\csc^2\theta}{\cot^2\theta} = \dfrac{\frac{1}{\sin^2\theta}}{\frac{\cos^2\theta}{\sin^2\theta}} $ This is $\dfrac{1}{\cos^2\theta} = \sec^2\theta$.